An object is launched at 29.4 meters per
second (m/s) from a 34.3-meter tall
platform. The equation for the object's
height s at time x seconds after launch is
f(x) = -4.9x2 + 29.4x+ 34.3, where y is in
meters. What is the initial height of the
object?

29.4 meters
78.4 meters
34.3 meters
3 meters

34.3 meters

Step-by-step explanation:

The generic equation for a movement with constant acceleration is:

S = So + Vo*t + (a*t^2)/2

Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time.

If we compare with our equation (where x is the time and f(x) is the final distance), we have that:

So = 34.3

Vo = 29.4

a = -9.8

So we have that the inicial position (So) of the object is 34.3 meters

-4

step-by-step explanation:

do 6-10=

if u need it to be negative.

answer:   y=2x-3

step-by-step explanation: